Answer
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Hint: Equivalent weight defined as the weight of an element or compound which combines with or displaces 8 grams of Oxygen or 1.00797 grams of Hydrogen.
Complete step by step answer:
> Phosphoric acid is an acid containing four atoms of Oxygen, one atom of Phosphorus and three atoms of hydrogen. It is tribasic acid.
IUPAC name of \[{{H}_{3}}P{{O}_{4}}\] is phosphoric acid which is also known as Phosphoric (V) or Orthophosphoric acid. It is a colourless white crystal like solid and weak acid which is used to make phosphate salts for fertilizers.
It is non -toxic and non -volatile. The widely used phosphoric acid concentration is 85% in \[{{H}_{2}}O\] water.
To calculate the equivalent weight of \[{{H}_{3}}P{{O}_{4}}\], First we will calculate the molecular weight of phosphoric acid.
> Calculation of molecular weight of Phosphoric acid:
As we know the atomic weight of the combining elements will give the molecular weight of the compound.
And we also know the atomic weight of \[H=1\], \[P=31\] and \[O=16\]
Therefore,
\[\Rightarrow 3(H)+1(P)+4(O)\]
\[\begin{align}
& \Rightarrow 3(1)+1(31)+4(16) \\
& \Rightarrow 3+31+64 \\
& \Rightarrow 98 \\
\end{align}\]
> Therefore from the above calculation, the molecular weight of phosphoric acid is 98 or it can also be written as M.
Now, based on the definition the formula of equivalent weight is represented as
\[\mathbf{Equivalent}\text{ }\mathbf{weight}\text{ }=\text{ }\mathbf{Molecular}\text{ }\mathbf{weight}/\text{ }\mathbf{Number}\text{ }\mathbf{of}\text{ }\mathbf{equivalent}\text{ }\mathbf{moles}\]
\[\Rightarrow ~Molecular\text{ }Mass\text{ }\left( M \right)/\text{ }Number\text{ }of\text{ }replaceable\text{ }hydrogens\]
\[\Rightarrow Number\text{ }of\text{ }replaceable\text{ }hydrogens\text{ }is\text{ }3\]
\[\Rightarrow Hence,\text{ }98/3\text{ }or\text{ }M/3\]
Hence, the correct option is (B).
Note: Don’t confuse equivalent weight with molecular weight of an element or compound because molecular weight of an element or compound which is also called as molecular mass defined as the mass of substance based on 12 as atomic weight of carbon is 12.
Complete step by step answer:
> Phosphoric acid is an acid containing four atoms of Oxygen, one atom of Phosphorus and three atoms of hydrogen. It is tribasic acid.
IUPAC name of \[{{H}_{3}}P{{O}_{4}}\] is phosphoric acid which is also known as Phosphoric (V) or Orthophosphoric acid. It is a colourless white crystal like solid and weak acid which is used to make phosphate salts for fertilizers.
It is non -toxic and non -volatile. The widely used phosphoric acid concentration is 85% in \[{{H}_{2}}O\] water.
To calculate the equivalent weight of \[{{H}_{3}}P{{O}_{4}}\], First we will calculate the molecular weight of phosphoric acid.
> Calculation of molecular weight of Phosphoric acid:
As we know the atomic weight of the combining elements will give the molecular weight of the compound.
And we also know the atomic weight of \[H=1\], \[P=31\] and \[O=16\]
Therefore,
\[\Rightarrow 3(H)+1(P)+4(O)\]
\[\begin{align}
& \Rightarrow 3(1)+1(31)+4(16) \\
& \Rightarrow 3+31+64 \\
& \Rightarrow 98 \\
\end{align}\]
> Therefore from the above calculation, the molecular weight of phosphoric acid is 98 or it can also be written as M.
Now, based on the definition the formula of equivalent weight is represented as
\[\mathbf{Equivalent}\text{ }\mathbf{weight}\text{ }=\text{ }\mathbf{Molecular}\text{ }\mathbf{weight}/\text{ }\mathbf{Number}\text{ }\mathbf{of}\text{ }\mathbf{equivalent}\text{ }\mathbf{moles}\]
\[\Rightarrow ~Molecular\text{ }Mass\text{ }\left( M \right)/\text{ }Number\text{ }of\text{ }replaceable\text{ }hydrogens\]
\[\Rightarrow Number\text{ }of\text{ }replaceable\text{ }hydrogens\text{ }is\text{ }3\]
\[\Rightarrow Hence,\text{ }98/3\text{ }or\text{ }M/3\]
Hence, the correct option is (B).
Note: Don’t confuse equivalent weight with molecular weight of an element or compound because molecular weight of an element or compound which is also called as molecular mass defined as the mass of substance based on 12 as atomic weight of carbon is 12.
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